This material is based upon work supported by the National Science Foundation under Grant No. DUE-0336493
STEP- Science, Technology, Engineering, and Mathematics Talent Expansion Program
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Using Set Notation, Domain, Variable Expressions, Evaluating Expressions, Simplifying Variable Expressions--Combining Like Terms, Simplifying Variable Expressions--Multiplication and the Distributive Property, Solving Equations, Translating, Word Problems
USING SET NOTATION
Answers to math problems are often expressed as elements of a set.
In the problem x + 4 = 9, the answer for the x can be written using set notation such as: x = {5} or {x|5}. The first example tells the reader that the answer for x is 5. The second example (set builder notation) is read as follows: the values for x such that x is equal to 5.
In the problem x² + 4 = 8, the values for x can be written as x = {-2,2} or {x | -2,2}. The second example (set builder notation) is read as follows: the values for x such that x is equal to -2 or 2.
Set-builder notation is helpful to describe infinite sets. For example {x | x is a quotient of two integers, with denominator not 0} describes the set of rational numbers. The information inside the braces is read as "the set of numbers x such that x is the quotient of two integers, with denominator not 0."
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DOMAIN
A variable is a symbol that is used to represent values for any element in a set. The domain is the set of numbers from which the values for the variable can be chosen.
For example: In the problem x + 7 = 10, the value for x (3) is selected from the domain that contains all Real numbers. The domain can be written as follows: {x | all Real numbers}. The only value from the set of Real numbers that will satisfy the given equation is x = {3}.
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VARIABLE EXPRESSIONS
Expressions do not contain equals signs. Variable expressions contain variables and numbers. The value of a variable expression changes depending on the value chosen to replace the variable or variables.
In the example 4n, the value for the expression is open until a value for the variable is chosen from the domain, replaced for the variable, and the multiplication is completed.
Examples of variable expressions are: 4x + 5; –2c; 7a–b; ⅓Y + 2; –6m – 10n; 8z; a + b + c
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EVALUATING EXPRESSIONS
The process for finding the number an expression represents is called evaluating the expression. To evaluate an expression, substitute the values for the variables and complete the problem according to the order of operations agreement.
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SIMPLIFYING VARIABLE EXPRESSIONS
COMBINING LIKE TERMS
3x² + 2xy – 5x + y – 8 + 9x² + 12 – 4y
Variable expressions contain terms. The terms are separated by plus and minus signs. There are eight terms in the variable expression above.
The terms can be either numerical or variable.
Numerical terms contains only numbers. The numerical terms in the variable expression above are: –8 and 12.
Variable terms contain variables and have two parts--the variable part and the numerical coefficient. The variable terms in the expression are: 3x², 2xy, –5x, y, 9x², and –4y.
The numerical coefficient is the number in front of the variable. Variable terms that appear to be only variables have a coefficient of one. The coefficients of the variable terms in the expression are: 3, 2, –5, 1, 9, and –4.
3x² + 2xy – 5x + y – 8 + 9x² + 12 – 4y
Like terms are terms that are identical in the variable part. Like terms have nothing to do with signs. Like terms have the same variables with the same exponents. There are three pairs of like terms in the above expression: 3x² and 9x², y and –4y, and –8 and 12. –5x is not like 3x² and 9x² because the exponent is different. –8 and 12 are like terms because they are both numerical terms.
Like terms may be combined by adding their coefficients using the addition rules.
The above variable expression can be simplified by combining like terms resulting in the following variable expression which contains 5 terms: 12x² + 2xy – 5x – 3y + 4.
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SIMPLIFYING VARIABLE EXPRESSIONS
MULTIPLICATION AND THE DISTRIBUTIVE PROPERTY
Addition is specific for like terms. However, like and unlike terms can be multiplied.
Use the distributive property to multiply addition problems by a numerical term or variable term.
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SOLVING EQUATIONS
Solving equations is a process that isolates the variable in an equation for the purpose of finding the value for the variable that balances or satisfies the equation. In the equation x + 2 = 8, isolating the variable results in the transformed equation x = 6, which is the value for x that balances the equation. Six is called the solution to the equation. The solution to an equation is often written in set notation as {6} or {x|6}.
The first step in solving an equation is to simplify the expressions on both sides of the equals sign. Follow the Order of Operations Agreement when simplifying the expressions. Remember to simplify using multiplication before simplifying using addition (combining like terms).
The second step in solving an equation is to transform the equation, using the addition property of equations.
The third step in solving an equation is to transform the equation, using the multiplication property of equations.
The fourth step in solving an equation is to check the solution by substituting the value calculated in the given equation for the variable in the original equation. A correct solution will satisfy both expressions on each side of the equals sign and balance the equation. Perform the checking process carefully. The arithmetic is usually more difficult in the process of checking the solution than in the solving process.
THREE EXAMPLES FOR SOLVING EQUATIONS
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TRANSLATING
Problem solving often requires translating phrases into mathematical symbols.
Translating Tips
Read and translate from left to right except for "less than" phrases.
Use common sense.
The word "and" does not mean addition. The word "and" is used to separate the objects in the mathematical operation.
"Is" indicates the equals sign.
Translating Examples
word or phrase example equation
sum of the sum of 2 and 3 is 5 2 + 3 = 5
total of the total of 3 and 4 is 7 3 + 4 = 7
more than 12 more than 4 is 16 12 + 4 = 16
increased by –4 increased by 13 is 9 –4 + 13 = 9
difference between the difference between 5 and 1 is 4 5 – 1 = 4
subtracted from 7 subtracted from 4 is –3 4 – 7 = –3
less 6 less 3 is 3 6 – 3 = 3
less than 6 less than 3 is –3 3 – 6 = –3
product of the product of 5 and 4 is 20 5 • 4 = 20
twice (multiply by 2) twice 6 is 12 2(6) = 12
of (used with fractions) ⅓ of 30 is 10 ⅓ (30) = 10
percent of 12% of 16 is 1.92 .12 (16) = 1.92
quotient of the quotient of –42 and 6 is –7 –42 ÷ 6 = –7
divided by 56 divided by –7 is 8 56 ÷ –7 = 8
ratio of the ratio of 3 to 4 ¾
More Translating Examples
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WORD PROBLEMS
In solving problems, the initial objective is to translate the English phrases into the language of algebra.
Consider the following examples:
If whole milk costs 25 cents more per gallon than skimmed milk, express the cost of a gallon of whole milk in terms of the cost of a gallon of skimmed milk.
Solution: Since the cost of skimmed milk could be any amount greater than zero, use a variable to represent the cost: let x = the cost of skimmed milk; therefore, x + 0.25 = the cost of whole milk expressed in dollars
Jerry has 6 less than one-third as many baseball cards as Bill. Write an algebraic expression for the number of Bill's baseball cards.
Solution: Since Bill could have any number of baseball cards, use a variable to represent the number of his baseball cards: let x = the number of Bill's cards, then ⅓x = one-third the number of Bill's baseball cards; therefore, ⅓x – 6 = the number of Jerry's baseball cards.
A chemistry teacher wants to build a shelf that will hold 10 reagent bottles. If each bottle has a diameter of d, how long should the shelf be?
Solution: 10d
Four Steps to Solving Word Problems
In "word problems" numbers are expressed in phrases and related to one another. To solve "word problems" determine the described numbers so that the relationships will be true.
Consider the following example:
The main body of a missile is 9 times as long as the nose cone. The entire missile is 100 feet in length. How long is the nose cone?
Solution:
STEP ONE--Label the Variable Choose a variable to represent the number you want to find and use the variable in describing all other numbers. Let x = length of the nose cone. Since the problem tells us that the main body is 9 times as long as the nose cone (the nose cone is labeled x), let 9x = length of the main body.
STEP TWO--Write the Equation Use the facts given in the problem to write the equation. Show that the sum of the two lengths labeled in step one are equal to 100.
length of nose cone + length of main body = 100 x + 9x = 100
STEP THREE--Solve the Equation x + 9x = 100 10x = 100 x = 10
STEP FOUR--Determine if the answer is correct. Use the words in the problem to determine if the answer is sensible. Length of the nose cone = 10, length of the main body = 9(10) = 90, total of the two lengths = 10 + 90 = 100. The answer is sensible and the math checks.
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